n个列表的Python3列表理解答案

【问题标题】：Python3 List Comprehension for n listsn个列表的Python3列表理解
【发布时间】：2020-05-30 22:02:39
【问题描述】：

我有一种方法可以按照我的意愿遍历我的列表，如下所示：

a = ["1","2","3","4","5","6","7","8","9","10"]
b = ['A','B','C','D','E','F','G','H','I','J']
c = ["11","12","13","14","15","16","17","18","19","20"]

for x, y, z in [(x,y,z) for x in a for y in b for z in c]:
    print(x,y,z)

输出：

但是，如果我的列表存储在一个列表中并且有 n 个列表，我怎样才能获得相同的结果？例如

main_list=[["1","2","3","4","5","6","7","8","9","1"],['A','B','C','D','E','F','G','H','I','J'],["11","12","13","14","15","16","17","18","19","20"],['k','l','m','n','o','p','q','r','s','t']]

提前致谢。

【问题讨论】：

看看zip function。本质上，如果main_list 包含所有列表，zip(*main_list) 应该会给你你想要的。
zip 不会这样做。 zip 只会给出“1 A 11, 2 B 12, 3 C 13, ...”

标签： python-3.x list list-comprehension

【解决方案1】：

看itertools.product：

a = ["1","2","3","4","5","6","7","8","9","10"]
b = ['A','B','C','D','E','F','G','H','I','J']
c = ["11","12","13","14","15","16","17","18","19","20"]

from itertools import product

all_lists = [a, b, c]
for c in product(*all_lists):
    print(c)

打印：

('1', 'A', '11')
('1', 'A', '12')
('1', 'A', '13')
('1', 'A', '14')
('1', 'A', '15')
('1', 'A', '16')
('1', 'A', '17')
('1', 'A', '18')
('1', 'A', '19')
('1', 'A', '20')
('1', 'B', '11')

... and so on.

【讨论】：

【解决方案2】：

例如，下面给出了所需的组合。

import itertools

list(itertools.product(*(a,b,c)))

对于主列表 -


main_list=[["1","2","3","4","5","6","7","8","9","1"],['A','B','C','D','E','F','G','H','I','J'],["11","12","13","14","15","16","17","18","19","20"],['k','l','m','n','o','p','q','r','s','t']]

combintns = list(itertools.product(*main_list))

# in case you're specific about output format/appearance
for i in combintns:
  print(i[0],i[1],i[2],i[3])

【讨论】：